Trigonometry - Previous Year IPM/BBA Questions
The best way to prepare for Trigonometry is by going through the previous year Trigonometry questions for IPMAT - Indore. Here we bring you all previous year Trigonometry IPMAT - Indore questions along with detailed solutions.
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If cosα + cosβ = 1, then the maximum value of sinα – sinβ is
- (a)
2
- (b)
√2
- (c)
1
- (d)
√3
Answer: Option D
Text Explanation :
Workspace:
If logcosxsinx + logsinxcosx = 2, then the value of x is
- (a)
nπ/4 + π/4, n is an integer
- (b)
2nπ + π/4, n is an integer
- (c)
nπ + π/4, n is an integer
- (d)
nπ/4, n is an integer
Answer: Option B
Text Explanation :
Workspace:
Sin α + Sin β = and Cos α + Cos β = , then the value of is _______.
Answer: 100
Text Explanation :
Sin α + Sin β = ...(1) and
Cos α + Cos β = ...(2),
Squaring both the equations and adding them we get.
⇒ Sin2 α + Sin2 β + 2 × Sin α × Sin β + Cos2 α + Cos2 β + 2 × Cos α × Cos β = +
⇒ Sin2 α + Cos2 α + Sin2 β + Cos2 β + 2 × Sin α × Sin β + 2 × Cos α × Cos β = 1
⇒ 1 + 1 + 2 × cos (α - β) = 1
⇒ cos (α - β) = - 1/2
⇒ 2 × - 1 = - 1/2
⇒ = 1/4
∴ 202 × = 400
⇒ = 100.
Hence, 100.
Workspace:
For 0 < θ < π/4, let a = ((sinθ)sinθ)(log2cosθ), b = ((cosθ)sinθ)(log2sinθ), c = ((sinθ)cosθ)(log2cosθ) and d = ((sinθ)sinθ)(log2sinθ). Then, the median value in the sequence a, b, c, d is
- (a)
(a + b)/2
- (b)
(a + d)/2
- (c)
(b + c)/2
- (d)
(c + d)/2
Answer: Option B
Text Explanation :
0 < θ < π/4, let's take θ = 30°
sin30° = 1/2 = 0.5, cos30° = √3/2 = 0.866
a = ((sinθ)sinθ)(log2cosθ),
⇒ a = ((1/2)1/2)(log20.866)
b = ((cosθ)sinθ)(log2sinθ),
⇒ b = ((0.866)1/2)(log21/2),
c = ((sinθ)cosθ)(log2cosθ)
⇒ c = ((1/2)0.866)(log20.866)
d = ((sinθ)sinθ)(log2sinθ)
⇒ d = ((1/2)1/2)(log21/2)
(1/2)0.5 > (1/2)0.866 and (log20.866) is negative
∴ c is greater than a.
Similarly, c > a > d > b.
∴ Median of these four numbers = average of 2 middle numbers = (a + d)/2
Hence, option (b).
Workspace:
The curve represented by the equation + = 1 is
- (a)
an ellipse with the foci on the y-axis
- (b)
an ellipse with the foci on the x-axis
- (c)
a hyperbola with the foci on the x-axis
- (d)
a hyperbola with the foci on the y-axis
Answer: Option A
Text Explanation :
Equation represents an ellipse.
Hence, the given curve is an ellipse. We need to figure out the foci of this ellipse.
Also, Cos√2 - cos√3 > Sin√2 - Sin√3 > 0
⇒ b2 > a2
∴ Foci is on y-axis.
Hence, option (a).
Workspace:
Ayesha is standing atop a vertical tower 200 m high and observes a car moving away from the tower on a straight, horizontal road from the foot of the tower. At 11:00 AM, she observes the angle of depression of the car to be 45°. At 11:02 AM, she observes the angle of depression of the car to be 30°. The speed at which the car is moving is approximately
- (a)
6.3 km per hour
- (b)
8.45 km per hour
- (c)
10.6 km per hour
- (d)
4.39 km per hour
Answer: Option D
Text Explanation :
At 11 am, the car is at C and Ayesha is at A.
In △ABC, angle ACB = 45°
Tan45° = 1 =
⇒ BC = AB = 200
At 11:02 am, the car is at D and Ayesha is at A.
In △ABD, angle ADB = 30°
Tan30° = =
⇒ BD = AB√3 = 200√3
⇒ CD = 200√3 - 200 = 200(√3 - 1) = 200(0.732) = 146.4 m
∴ The car moved 146.4 m in 2 minutes.
⇒ Speed of the car = = 4.392 kmph.
Hence, option (d).
Workspace:
The set of all real value of p for which the equation 3sin2x + 12cosx – 3 = p has at least one solution is
- (a)
[-12, 12]
- (b)
[-12, 9]
- (c)
[-15, 9]
- (d)
[-15, 12]
Answer: Option C
Text Explanation :
The set of all real value of p for which the equation 3sin2x + 12cosx – 3 = p has one solution is
Given, 3sin2x + 12cosx – 3 = p
⇒ 3(1 - cos2x) + 12cosx – 3 = p
⇒ -3cos2x + 12cosx – p = 0
⇒ -3[cos2x - 4cosx] – p = 0
⇒ -3[cos2x - 4cosx + 4] + 12 – p = 0 [Adding and subtracting 12]
⇒ -3[cosx - 2]2 = p - 12
⇒ [cosx - 2]2 = -p/3 + 4
Now, 1 ≥ cosx ≥ -1
∴ 9 ≥ (cosx - 2)2 ≥ 1
⇒ 9 ≥ -p/3 + 4 ≥ 1
⇒ 5 ≥ -p/3 ≥ -3
⇒ -15 ≤ p ≤ 9
Hence, option (c).
Workspace:
The value of + + +
- (a)
1
- (b)
3/2
- (c)
2
- (d)
9/4
Answer: Option C
Text Explanation :
Given, + + +
= + + +
= + + +
= + + + +
= 1 + 1 = 2
Hence, option (c).
Workspace:
The number of pairs (x, y) satisfying the equation sinx + siny = sin(x + y) and |x| + |y| = 1 is
Answer: 6
Text Explanation :
Workspace:
Given that cos x + cos y = 1, the range of sin x - sin y is
- (a)
[-1, 1]
- (b)
[-2, 2]
- (c)
[0, √3]
- (d)
[-√3, √3]
Answer: Option D
Text Explanation :
Workspace:
If sinθ + cosθ = m, then sin6θ + cos6θ equals
- (a)
3(m2 + 1)/4
- (b)
3(m2 - 1)/4
- (c)
1 - 3(m2 - 1)/4
- (d)
1 - 3(m2 - 1)2/4
Answer: Option D
Text Explanation :
Given, sinθ + cosθ = m
Taking square boths sides we get
sin2θ + cos2θ + 2 × sinθ × cosθ = m2
⇒ 1 + 2 × sinθ × cosθ = m2
⇒ sinθ × cosθ = (m2 - 1)/2 ...(1)
Now we need to find the value of sin6θ + cos6θ
= (sin2θ)3 + (cos2θ)3
= (sin2θ + cos2θ)(sin4θ + cos4θ - sin2θ × cos2θ)
= 1 × ((sin2θ + + cos2θ)2 - 2 × sin2θ × cos2θ - sin2θ × cos2θ)
= ((1)2 - 3 × sin2θ × cos2θ)
= (1 - 3 × sin2θ × cos2θ)
= (1 - 3 × ((m2 - 1)/2)2)
= (1 - 3 × (m2 - 1)2/4)
Hence, option (d).
Workspace: